Question

If \(7\hat{j}+10\hat{k},-\hat{i}+6\hat{j}+6\hat{k}\) and \(-4\hat{i}+9\hat{j}+6\hat{k}\) are the position vectors of the vertices A, B and C repectively of \(△ABC\). Then the position vector of the point where the bisector of the angle A meets side BC is

• A. \((2+3\sqrt{2})\hat{i}+(3+3\sqrt{3})\hat{j}+6\hat{k}\)
• B. \((2-3\sqrt{2})\hat{i}+(3+3\sqrt{2})\hat{j}+6\hat{k}\)
• C. \((2-3\sqrt{2})\hat{i}+(3-3\sqrt{2})\hat{j}+6\hat{k}\)
• D. \((2-3\sqrt{2})\hat{i}+(3+3\sqrt{2})\hat{j}-6\hat{k}\)

Hint:

The bisector divides $BC$ in the ratio $AB:AC$.

Answer 1

The Correct Option is B

Step 1: Find the side lengths
\(A(0,7,10)\), \(B(-1,6,6)\), \(C(-4,9,6)\).
\(AB=\sqrt{1+1+16}=3\sqrt2\) and \(AC=\sqrt{16+4+16}=6\).

Step 2: Use the angle bisector theorem
\(BD:DC=AB:AC=3\sqrt2:6=\sqrt2:2\).

Step 3: Section formula
\[ D=\frac{2B+\sqrt2C}{2+\sqrt2} \]
Numerator: \((-2-4\sqrt2,\;12+9\sqrt2,\;12+6\sqrt2)\). Multiply by \(\frac{2-\sqrt2}{2}\) to rationalise:
\(x=\frac{(-2-4\sqrt2)(2-\sqrt2)}{2}=2-3\sqrt2\), \(y=\frac{(12+9\sqrt2)(2-\sqrt2)}{2}=3+3\sqrt2\), \(z=\frac{(12+6\sqrt2)(2-\sqrt2)}{2}=6\).

Final Answer:
The point is \((2-3\sqrt2)\hat i+(3+3\sqrt2)\hat j+6\hat k\), option (B). \[ \boxed{\text{(B)}} \]